Sunday, December 27, 2015

Problem : We need to prove that any isosceles trapezium is con-cyclic.

Solution :Here’s an isosceles trapezium with sides PQ and RS are parallel and PS = QR. A quadrilateral is cyclic if it’s opposite angles are supplementary (i.e. they add up to 180˚). So , we need to prove that QPS + QRS = 180˚ and PSR + PQR = 180˚.

Let us construct two perpendiculars, PU and QT, from point P and Q to segment RS.

Monday, December 21, 2015

Problem : Let us consider the right triangle PQR with the right angle P (Figure 1), and let PS be the median drawn from the vertex P to the hypotenuse QR. We need to find the relationship between the length of the median PS and the the length of the hypotenuse QR.

Solution :Draw a straight line passing through the midpoint S and parallel to the side PR intersecting the side PQ at the point T. (Figure 2).

The angle QPR is given as right angle. The angles QTS and QPR are equal to each other as as they are corresponding angles of the parallel lines PR and TS and the transversal PQ. Hence the angle QTS is a right angle.

As TS passes through the mid-point S and is parallel to PR , it divides the side PQ into two equal parts i.e. PT = TQ. So, the triangles PTS and QTS are right triangle triangles with equal sides PT and TQ , these triangles also have a common side TS. Hence, these triangles are congruent in as per the Side – Angle – Side (SAS) Rule.

From this we can say that the other sides of these triangles are also equal to each other as they are the corresponding parts of the congruent triangles , thus PS = QS. Now QS is equal to half the length of the hypotenuse QR , we can say that the median PS is also equal to half the length of the hypotenuse.

Hence, we can conclude that in a right triangle , the length of median to hypotenuse is half the length of the hypotenuse.

Friday, December 18, 2015

Locus Problem : A ladder of length L is sliding over the floor. Find the locus of point D such that D is the foot of perpendicular dropped from point C such that OACB is a rectangle.

Solution: Let A (a,0) be the feet of the ladder and B (0,b) be the top of the ladder rest against the wall . O is the intersection point of floor and wall. OACB is a rectangle. D is the feet of the perpendicular drawn from point C onto ladder AB . As the feet of the ladder slides, the point D travels on a path called asteroid whose equation is given by x(2/3)+y(2/3)=L(2/3).

In the following applet, press the play button at the lower left corner to see the various positions of the ladder as it is dragged and the movement of point D of the ladder. You can also select the ‘Show Locus’ button to see the locus and its equation.

Wednesday, December 16, 2015

In continuation to my previous post on locus of mid-point of a falling ladder , let us now explore the locus of a point lying anywhere on the falling ladder.

Example : A 6-foot ladder is placed vertically against a wall, and then the foot of the ladder is moved outward until the ladder lies flat on the floor with one end touching the wall. What is the locus of the point which divides the ladder in the ratio 2 : 3 as it slides?

Solution: Let P be the point which divides the slider AB (A (a,0) being the feet and B (0,b) rest on the wall) .O is the intersection point of floor and wall.As the feet of the ladder slides,the point P whose coordinates are(3a/5,2b/5) travels on an elliptical path given by x2/9+y2/4=(6/5)2.

In the following applet, press the play button at the lower left corner to see the various positions of the ladder as it is dragged and the movement of point P of the ladder. You can also select the ‘Show Locus’ button to see the full locus and its equation.